# Work done by friction on a sphere sliding down the inclined plane [closed]

If a sphere slides down from rest on an inclined plane of hypotenuse $$L$$ and height $$H$$ : Work done by friction is converted into rolling kinetic energy and as well as heat energy.

If $$f$$ is the friction force then is work done by friction $$fL =$$Heat energy $$+$$ rolling $$KE$$ ? Or does the formula $$fL$$ gives only the heat energy lost by friction force ?

My second question regarding this is : Is it correct to write work- energy theorem at the feet of the inclined plane as :

$$KE\ _{linear}=mgh+fl$$ This not as banal a problem as you may expect at first sight.

First, study the emerging rotational motion:

$$F_N=mg\cos\theta$$ $$F_f=\mu_k F_N=\mu_k mg\cos\theta$$ Torque about the axis of rotation causes angular acceleration: $$\tau=I\alpha$$ $$F_f R=I\frac{\text{d}\omega}{\text{d}t}$$ $$\mu_k mg\cos\theta R=\gamma mR^2 \frac{\text{d}\omega}{\text{d}t}$$ where $$\gamma$$ is a coefficient depending on the exact shape of the rotating body. $$\frac{\text{d}\omega}{\text{d}t}=\frac{\mu_kg\cos\theta}{\gamma R}$$ Assuming $$\omega =0$$ at $$t=0$$: $$\omega(t)=\frac{\mu_kg\cos\theta}{\gamma R}t$$ Now study the translational motion: $$F_s-F_f=ma$$ $$mg\sin\theta-\mu_k mg\cos\theta=ma$$ $$\frac{\text{d}v}{\text{d}t}=g(\sin\theta-\mu_k \cos\theta)$$ Assuming $$v=0$$ at $$t=0$$: $$v(t)=g(\sin\theta-\mu_k \cos\theta)t$$ The object reaches rolling without slipping (pure rolling) when:

$$v(t)=\omega(t)R$$

which with some substituting and reworking gives the relationship:

$$\mu_k=\frac{\gamma}{\gamma+1}\tan\theta$$

So how to calculate the relevant energies?

You already know the work done by the friction force.

How much energy is used to get the object to roll? Calculate the time needed to reach the bottom of the incline ($$0 \to L$$) and from there calculate $$\omega(t)$$ and use that to calculate the change in rotational kinetic energy.

I hope this helps.

• Can you please explain if The sphere is only sliding during the motion , that is $v \ne \omega r$ and i use the formula Work done by friction $fL$ , it means the total work done by friction(that's what i think you mean) or just the heat loss by frictional force ? Aug 3, 2020 at 18:01
• It's only sliding when $v \neq \omega R$, correct. After that, during pure rolling, $F_f$ no longer does work. That's why you need to determine the point $t$ where $v=\omega R$.
– Gert
Aug 3, 2020 at 18:11
• Sir I am only talking about the case when during the whole motion the sphere slides and never enters pure rolling motion. Aug 3, 2020 at 18:14
• Well, that's fine, no problemo! In that case calculate all the energies and know that the total energy is the change in potential energy: $\Delta U=\Delta E_{trans}+\Delta E_{rot}+\Delta Q_{heat}$.
– Gert
Aug 3, 2020 at 18:18
• And can we write $\Delta E_{rot} + \Delta Q_{heat}=fl$? I just want to be absolutely sure. Aug 3, 2020 at 18:22

If you are considering kinetic energy only due to linear motion . The your equation is wrong . It's : $${KE\ _{linear}} + fl = mgh$$

Because friction is working against the linear motion.

• I am also considering rotational kinetic energy. Aug 3, 2020 at 15:43
• You wrote linear in subscript Aug 3, 2020 at 15:45
• I specified linear motion because I don't know how to write in subscript. Pls tell Aug 3, 2020 at 15:46
• I wrote linear KE in the subscript doesn't mean that i forget rotational KE. Aug 3, 2020 at 15:47
• I edited your answer. You can see what changes i have made to learn how to write in subscript. Aug 3, 2020 at 15:50